The young Meisami Mahdi sent the following
Conjectures closely related to the well known Gilbreath scheme &
conjecture. (See
1
& our
puzzle 274)

For a function f from 1 to N define
D(f)(n)=f(n+1)-f(n) as a derivative.

For a list {f(1),f(2),f(3),.....} of values of f define:

D0(f):={f(1),f(2),f(3),.....}
D1(f):={D(f)(1),D(f)(2),D(f)(3),......}=D(
{f(1),f(2),f(3),.....} )
D2(f):={D(D1(f))(1),D(D2(f))(2),....}=D( {D(f)(1),D(f)(2),D(f)(3),......}
) and so on for Dk(f) for every k.

As an example consider f(n)=3n+1 or {4,7,10,13,16,....},
D1(f)={3,3,3,3,3,3,....}.
Another example take g(n)=n^2 or {1,4,9,16,25,36,49,....}
D1(g)={3,5,7,9,11,13,....} , D2(g)={2,2,2,2,2,2,2,....}

Conjecture1: Let p(n) be prime number function, then
there is not any k for which all elements of the set Dk(p)
are equal.

Conjecture 2: (a conjecture stronger than
conjecture1): There is a m such that for every k greater
than m,Dk(p) does not have any two equal elements.

Conjecture 3: For every natural number m there is a k
such that m is an element of Dk(p).

But if we change the definition of D(f)(n) and define a new
derivative AD(f)(n)=|f(n+1)-f(n)| (=Absolute value of
D(f)(n)) like something we have in Gilbreath's conjecture
(the first element of each ADk(p) is 1).

Conjecture 4:There is a m such that for every k
greater than m we have just 1,0,2 as elements of ADk(p).

Q. Can you prove the
conjectures by Mahdi or show a counterexample?

Contributions came from Jan van Delden & Emmanuel Vantieghem.

***

Jan wrote:

Conjecture 1: Consider g[] a polynomial with degree k. Then
D[k]g[] is constant, i.e. D[k]g[l]=c for every l.
Or the other way around, if there exists a k such that D[k]g[]
is constant (and D[m]g[] is not constant with g<k), we must
have that g[] is a polynomial with degree k.

So the conjecture is equivalent to p[] is not a polynomial
of degree k.

One can do a little better:

p[n] is integer, so all D[k]p[n] are integer by
construction. So if D[k]p[n] is constant it is an integer.
Reversing the process shows that we must have that p[] is a
polynomial with rational coefficients.

The conjecture is equal to: p[] is not a polynomial with
finite degree and rational coefficents.

This has been proved, in fact we have the stronger theorem
by Legendre:
There is norationalalgebraic
function which always gives primes.

Conjecture 2: Clearly m>1, since there are a lot of twin
primes. The conjecture states that the distribution of
primes is highly irregular. I agree, but proving this in
this form would be something.

[B.T.W.: SeeYitang
Zhangon a
proof that there exists a number N<70*10^6 such that there
are infinitely many primegaps equal to N.
There was an article in a Dutch Newspaper regarding his
remarkable proof.]

Conjecture 3: A bold statement. And hard to prove or
disprove.

Conjecture 4: I investigated M[N], where M is the first
value of k for which AD[k]p[] is in {0,1,2}, with p[]
restricted to [1,N]. So slightly different from m.

N M[N]

1 0

5 2

11 3

18 4

19 5

31 9

35 10

70 11

110 13

112 15

190 16

193 17

194 18

195 19

196 20

312 23

764 24

899 27

901 35

2216 39

See Caldwell’s pages for a table of M[pi(10^k)], i.e.
N=pi(10^k), k=2..13.

Curve fitting these values gives: M=0.393*ln(N)^2.249 with
R^2=0.998.

We see that M is a nondecreasing stepfunction of N. I see no
reason why M (and thus m) is bounded above by a constant.

***

Emmmanuel wrote:

About conjecture 1 : it is true and easy to prove. Indeed, one
sees immediately that, for every k > 0 the sequence Dk(p) has
first element odd and all other elements even. So, there is not any
k for which all elements of the sequence are equal.

About conjecture 3 : I'm convinced that it is not true. For
instance, 15 is not an element of any Dk(p) for k <
30000. The
first element of Dk(p) is the only odd number in Dk(p) and is
very big (in absolute value) in comparison with k .

About conjecture 4 : in the article of Wikipedia we read that
Odlyzko figured out that AD635(p) has only 0, 1 and 2 among his
first 3.4*10^11 elements. I think that it will be very difficult
to prove or disprove that ALL its elements are 0,1 or 2. In my
opinion the conjecture is not true.

It is clear that conjecture 4 implies the
Gilbreath conjecture. Indeed, if such an m exists then Dk(p)
would look like 1, followed by a sequence of zeros and twos. And
thus, the 1 in front would stay ad infinitum ...